You have found the following ages (in years) of all 5 lions at your local zoo: $ 4,\enspace 12,\enspace 6,\enspace 3,\enspace 6$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 5 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{4 + 12 + 6 + 3 + 6}{{5}} = {6.2\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-2.2$ years $4.84$ years $^2$ $12$ years $5.8$ years $33.64$ years $^2$ $6$ years $-0.2$ years $0.04$ years $^2$ $3$ years $-3.2$ years $10.24$ years $^2$ $6$ years $-0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4.84} + {33.64} + {0.04} + {10.24} + {0.04}} {{5}} $ $ {\sigma^2} = \dfrac{{48.8}}{{5}} = {9.76\text{ years}^2} $ The average lion at the zoo is 6.2 years old. The population variance is 9.76 years $^2$.